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NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

Edited By Ramraj Saini | Updated on Nov 17, 2023 05:21 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2

NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.2 contains two theorems: (Theorem 1) When a parallel line is drawn along one of the triangle's sides and intersects the other two sides at certain points, the two sides are divided in the same proportion. (Theorem 2) Any line that divides the two sides of a triangle in the same ratio is parallel to the third side.

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  1. NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 6 exercise 6.2
  3. Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.2
  4. More About NCERT Solutions for Class 10 Maths Exercise 6.2
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 6.2
  6. NCERT Solutions of Class 10 Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles
NCERT Solutions for Exercise 6.2 Class 10 Maths Chapter 6 - Triangles

NCERT solutions for exercise 6.2 Class 10 Maths chapter 6 Triangles discusses about the concept triangle as a polygon, interior angles of a triangle, exterior angles of triangle, Pythagoras' theorem, and conditions of similarity in triangles. 10th class Maths exercise 6.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.2

Triangles Class 10 Chapter 6 Exercise: 6.2

Q1 In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

1635923248436

Answer:

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

\frac{AD}{DB}=\frac{AE}{EC}

\Rightarrow \frac{1.5}{3}=\frac{1}{x}

\Rightarrow x=\frac{3}{1.5}=2\, cm

\therefore EC=2\, cm

(ii)

Let AD be x

Given: DE || BC

By using the proportionality theorem, we get

\frac{AD}{DB}=\frac{AE}{EC}

\Rightarrow \frac{x}{7.2}=\frac{1.8}{5.4}

\Rightarrow x=\frac{7.2}{3}=2.4\, cm

\therefore AD=2.4\, cm

Q2 (1) E and F are points on the sides PQ and PR respectively of a \triangle PQR. For each of the following cases, state whether EF || QR: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Answer:

(i)

1635923264991

Given :

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

\frac{PE}{EQ}=\frac{3.9}{3}=1.3\, cm and \frac{PF}{FR}=\frac{3.6}{2.4}=1.5\, cm

We have

\frac{PE}{EQ} \neq \frac{PF}{FR}

Hence, EF is not parallel to QR.

Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that \frac{AM}{AB} = \frac{AN}{AD }

1635923547862

Answer:

Given : LM || CB and LN || CD

To prove :

\frac{AM}{AB} = \frac{AN}{AD }

Since , LM || CB so we have

\frac{AM}{AB}=\frac{AL}{AC}.............................................1

Also, LN || CD

\frac{AL}{AC}=\frac{AN}{AD}.............................................2


From equation 1 and 2, we have

\frac{AM}{AB} = \frac{AN}{AD }

Hence proved.

Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC

1635923629376

Answer:


Given : DE || AC and DF || AE.

To prove :

\frac{BF}{FE} = \frac{BE}{EC }

Since , DE || AC so we have

\frac{BD}{DA}=\frac{BE}{EC}.............................................1

Also,DF || AE

\frac{BD}{DA}=\frac{BF}{FE}.............................................2

From equation 1 and 2, we have

\frac{BF}{FE} = \frac{BE}{EC }

Hence proved.

Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

1635923658934

Answer:


Given : DE || OQ and DF || OR.

To prove EF || QR.

Since DE || OQ so we have

\frac{PE}{EQ}=\frac{PD}{DO}.............................................1

Also, DF || OR

\frac{PF}{FR}=\frac{PD}{DO}.............................................2

From equation 1 and 2, we have

\frac{PE}{EQ} = \frac{PF}{FR }

Thus, EF || QR. (converse of basic proportionality theorem)

Hence proved.

Q6 In Fig. 6.21, A, B, and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

1635923722792

Answer:

Given : AB || PQ and AC || PR

To prove: BC || QR

Since, AB || PQ so we have

\frac{OA}{AP}=\frac{OB}{BQ}.............................................1

Also, AC || PR

\frac{OA}{AP}=\frac{OC}{CR}.............................................2

From equation 1 and 2, we have

\frac{OB}{BQ} = \frac{OC}{CR }

Therefore, BC || QR. (converse basic proportionality theorem)

Hence proved.

Q7 Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

1635923725135

Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.

i.e. PQ||BC and AP=PB .

Using basic proportionality theorem, we have

\frac{AP}{PB}=\frac{AQ}{QC}..........................1

Since AP=PB

\frac{AQ}{QC}=\frac{1}{1}

\Rightarrow AQ=QC

\therefore Q is the midpoint of AC.

Q8 Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer:

1635923738877

Let P is the midpoint of line AB and Q is the midpoint of line AC.

PQ is the line joining midpoints P and Q of line AB and AC, respectively.

i.e. AQ=QC and AP=PB .

we have,

\frac{AP}{PB}=\frac{1}{1}..........................1

\frac{AQ}{QC}=\frac{1}{1}...................................2

From equation 1 and 2, we get

\frac{AQ}{QC}=\frac{AP}{PB}

\therefore By basic proportionality theorem, we have PQ||BC

Q9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
that \frac{AO}{BO} = \frac{CO}{DO}

Answer:

1635923757337

Draw a line EF passing through point O such that EO||CD\, \, and\, \, FO||CD

To prove :

\frac{AO}{BO} = \frac{CO}{DO}

In \triangle ADC , we have CD||EO

So, by using basic proportionality theorem,

\frac{AE}{ED}=\frac{AO}{OC}........................................1

In \triangle ABD , we have AB||EO

So, by using basic proportionality theorem,

\frac{DE}{EA}=\frac{OD}{BO}........................................2

Using equation 1 and 2, we get

\frac{AO}{OC}=\frac{BO}{OD}

\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}

Hence proved.

Q10 The diagonals of a quadrilateral ABCD intersect each other at point O such that \frac{AO}{BO} = \frac{CO}{DO} Show that ABCD is a trapezium.

Answer:

1635923771348

Draw a line EF passing through point O such that EO||AB

Given :

\frac{AO}{BO} = \frac{CO}{DO}

In \triangle ABD , we have AB||EO

So, by using basic proportionality theorem,

\frac{AE}{ED}=\frac{BO}{DO}........................................1

However, its is given that

\frac{AO}{CO} = \frac{BO}{DO}..............................2

Using equation 1 and 2 , we get

\frac{AE}{ED}=\frac{AO}{CO}

\Rightarrow EO||CD (By basic proportionality theorem)

\Rightarrow AB||EO||CD

\Rightarrow AB||CD

Therefore, ABCD is a trapezium.

More About NCERT Solutions for Class 10 Maths Exercise 6.2

NCERT solutions Class 10 Maths chapter 6 exercise 6.2 triangles necessitates a thorough understanding of the underlying theorems. As a result, students are advised to read through the logical explanations of the theorems' proofs, as well as practise them using well-explained diagrams.

NCERT solutions Class 10 Maths chapter 6 exercise 6.2 may be mastered if students remember and comprehend how the theorems are applied. Children should consider creating a theory chart that they can refer to from time to time. Another creative and simple way to comprehend theorems is to use everyday objects. Students might ask their teachers and parents for assistance in doing the same. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.

Benefits of NCERT Solutions for Class 10 Maths Exercise 6.2

  • The NCERT solutions for Class 10 Maths exercise 6.2 guide students through the process of resolving and revising all of the exercise's questions.
  • You will be able to achieve better scores if you look over the NCERT solution for class 10 Mathematics chapter 6 exercise 6.2 and practise it thoroughly.
  • Congruent and related triangles and theorems are used in Class 10 Maths exercise 6.2.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What are the figures that are similar?

The term "similar" refers to two figures that have the same shape but differ in size.

2. Difference between congruent and similar figures?

The figures that are congruent are all similar, but the opposite is not true.


3. What is the use of side ratios for similar triangle?

If we know the side ratio and the length of the any side, we can find the length of the corresponding side of the other triangle by using the ratio.


4. What is the ratio of the area of triangles if the ratio of the sides is given?

The ration of area of the triangles is equal to the square of the ratio of the sides.


5. If the ratio of the sides of two similar triangle is 1:2 then the ratio of their areas is?

The ration of area of the triangles is equal to the square of the ratio of the sides.

Ratio of area = 1:4


6. If the ratio of the sides is given as 1:3 and the shorter side has length of 4cm then the length of the longer side is?

4/x=1/3

x=12cm

The length of the longer side is 12cm.


7. Mention one of the properties of the line that passes through the mid-point of any two sides of the triangle?

It will always be parallel to the third side of the triangle. 


8. Is similarity a sign of agreement between the figures?

No, resemblance only relates to the shape of the figures and does not imply equality.

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